# Recitation 14

Today I finished the proof for the bijectivity for the function $f: (0,1)\to\mathbb{R}$ defined by $$f(x)=\frac{2x-1}{2x(1-x)}$$. Once we choose a real number $a$, we shall show $f(x)=a$ has a unique solution in $(0,1)$. This equation is equivalent to $$2ax^2+(2-2a)x-1=0.$$

If $a=0$, there is only one solution $x=1/2$.

Otherwise $a\neq 0$, then we have two possible solutions, $$x_1 = \frac{a-1+\sqrt{a^2+1}}{2a}$$ and $$x_2=\frac{a-1-\sqrt{a^2+1}}{2a}.$$

By the trick of multiplying conjugate of the numerator, for instance $$x_1=\frac{(a-1)^2-(a^2+1)}{2a(a-1-\sqrt{a^2+1})}=\frac{1}{1+(\sqrt{a^2+1}-a)},$$ we could have an estimate on $x_1$. Here, since $\sqrt{a^2+1}-a > 0$, $x_1\in(0,1)$.

Similarly, one could show that $x_2\notin(0,1)$.